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Re: Re: Simplify with NestedLessLess?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg106581] Re: [mg106531] Re: [mg106487] Simplify with NestedLessLess?
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Sun, 17 Jan 2010 07:10:53 -0500 (EST)
  • References: <201001141049.FAA19892@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

You said C<<Cf, so let's say C = delta CF, where delta is small. We  
compute a series around delta=0 and, after giving it a look, we set delta  
= 0.

expr =
  Normal@Series[-Cf^2 L2^2 Rg^2 Vg^4 +
      3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /.
     C -> delta Cf, {delta, 0, 6}]
firstTry = expr /. delta -> 0

12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4 +
  delta (12 Cf Rg^2 Vd^2 + 6 Cf Rg^2 Vd Vg)

12 Cf Rg^2 Vd^2 - Cf^2 L2^2 Rg^2 Vg^4

In a more complicated situation (if a few Series terms were unconvincing),  
we might use Limit, instead:

expr = -Cf^2 L2^2 Rg^2 Vg^4 +
    3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg) /. C -> delta Cf
secondTry = Limit[expr, delta -> 0]

-Cf^2 L2^2 Rg^2 Vg^4 +
  3 (4 Cf Rg^2 Vd^2 + 4 Cf delta Rg^2 Vd^2 + 2 Cf delta Rg^2 Vd Vg)

Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4)

In this case, both answers are the same:

firstTry == secondTry // Simplify

True

Bobby

On Sat, 16 Jan 2010 13:38:22 -0600, Dave Bird <dbird at ieee.org> wrote:

> Interesting! But, I don't think I am correctly communicating what I'm  
> after yet. (Although, I admit that I am struggling some to keep up with  
> you guys in your Mathematica replies due to my inexperience.)
>
> The original expression that I put up for illustration is:
>
> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg)
>
> We compare 4 C Rg^2 Vd^2 to 4 Cf Rg^2 Vd^2 because the two terms share  
> common coefficients so that they "reduce" to (4 Rg^2 Vd^2+4 Rg^2 Vd^2)  
> (C+Cf) . Thus it becomes obvious that C may be discarded w.r.t. Cf.
>
> Please forgive if I have missed the correct application of your  
> suggestion, and thanks for the interest.
>
> Dave
>
> DrMajorBob wrote:
>> Series[-Cf^2 L2^2 Rg^2 Vg^4 +
>>    3 (4 C Rg^2 Vd^2 + 4 Cf Rg^2 Vd^2 + 2 C Rg^2 Vd Vg), {C, 0,
>>    5}] // Simplify
>>
>> SeriesData[C, 0, {
>>  Cf Rg^2 (12 Vd^2 - Cf L2^2 Vg^4), 6 Rg^2 Vd (2 Vd + Vg)}, 0, 6, 1]
>>
>> Bobby
>>
>> On Fri, 15 Jan 2010 02:21:09 -0600, Dave Bird <dbird at ieee.org> wrote:
>>
>>> Not infinitesimals. I'm working in analog circuit design/analysis. I
>>> have a 3 pole symbolic circuit response (third order) which is not
>>> easily separable. I can use Mathematica to find the three roots of the
>>> response. But, the roots are, of course, very messy. I know that  
>>> certain
>>> elements in the circuit are orders of magnitude larger than other like
>>> elements - capacitors in this case. For example, one small section of
>>> one root is
>>>
>>> -Cf^2 L2^2 Rg^2 Vg^4+3 (4 C Rg^2 Vd^2+4 Cf Rg^2 Vd^2+2 C Rg^2 Vd Vg)
>>>
>>> I know that C<<Cf. By careful inspection, I can see that the first term
>>> in the parens will drop out compared to the second term in the parens.  
>>> I
>>> would like Mathematica to do this without my having to examine it so
>>> closely since there are many other like situations.
>>>
>>> This kind of situation occurs in many other engineering situations.
>>>
>>> Hope this helps clarify.
>>>
>>> Thanks for the interest.
>>>
>>> Dave
>>>
>>>
>>>
>>>
>>> Daniel Lichtblau wrote:
>>>> Dave Bird wrote:
>>>>> Thanks Daniel for the observation. I forgot to add that both a, and b
>>>>> are real positive. That, of course would have to be added to the
>>>>> assumptions.
>>>>>
>>>>> Dave
>>>>
>>>> It's still not obvious what you are wanting to do. I have the idea you
>>>> are working in some sense with infinitesmals. If so, I doubt Simplify
>>>> would be the best tool for removing them; it really can only do that
>>>> if it is told, in some way, to replace them with zero. How might one
>>>> instruct Simplify to figure that out?
>>>>
>>>> Daniel
>>>>
>>>>
>>>>> Daniel Lichtblau wrote:
>>>>>> dbird wrote:
>>>>>>> Please excuse if this has been answered before, but I can't find  
>>>>>>> it.
>>>>>>>
>>>>>>> Is there some way to do a Simplify with assumptions using a
>>>>>>> NestedLessLess or something similar? For example:
>>>>>>>
>>>>>>> d=a+b
>>>>>>> Simplify[d,NestedLessLess[a,b]]
>>>>>>>
>>>>>>> Answer is:
>>>>>>> a+b
>>>>>>>
>>>>>>> Answer should be:
>>>>>>> b
>>>>>>>
>>>>>>> Thanks,
>>>>>>>
>>>>>>> Dave
>>>>>>
>>>>>> I fail to see why the result should be b.
>>>>>>
>>>>>> Daniel Lichtblau
>>>>>> Wolfram Research
>>>>>>
>>>>>>
>>>>
>>>>
>>>
>>
>>


-- 
DrMajorBob at yahoo.com


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